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Find the limiting reagent if 85.6 grams of Al were reacted with 498 grams of MnO.
2Al+3MnO=>Al2O3+3Mn
First we find the GFW of each reactant.
GFW of Al: 27
GFW of MnO: 71
Next we divide the number of grams of each reactant by the GFW.
85.6/27= 3.17
498/71= 7.01
Finally, we multiply each of the results by the coefficient of the other reactant/
3.17 x 3= 9.51
7.01 x 2= 14.02
Because 9.51 is smaller than 14.02, we know that the limiting reagent is Al.
2Al+3MnO=>Al2O3+3Mn
First we find the GFW of each reactant.
GFW of Al: 27
GFW of MnO: 71
Next we divide the number of grams of each reactant by the GFW.
85.6/27= 3.17
498/71= 7.01
Finally, we multiply each of the results by the coefficient of the other reactant/
3.17 x 3= 9.51
7.01 x 2= 14.02
Because 9.51 is smaller than 14.02, we know that the limiting reagent is Al.